A metal M has an atomic mass of 60. If atomic radius of metal is 7.53A and density of metal =1135 gm/ml, then what will be the edge length of the cubic crystal unit cell obtained using metal M?Express answer in nm NA=6×1023

Actual volume occupied by molecular of 1 mole=43πr3×NA1135gm atom occupies = 1 ml60 gm atom occupies =135×60 ml Efficiency =43πr3×NA135×60=43π×7.53×10−83×1023135×60=68%⇒unit cell is body centered3a=4ra=4×7.5×33=30 Å=3 nm

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A metal M has an atomic mass of 60. If atomic radius of metal is $7.5 \sqrt{3} A$ and density of metal $= \frac{1}{135} gm / ml$ , then what will be the edge length of the cubic crystal unit cell obtained using metal M?
Express answer in nm $[N_{A} = 6 \times 10^{23}]$

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Detailed Solution

Actual volume occupied by molecular of 1 mole

$= \frac{4}{3} π r^{3} \times N_{A}$

$\frac{1}{135}$ gm atom occupies = 1 ml

60 gm atom occupies $= 135 \times 60 ml$

$Efficiency = \frac{\frac{4}{3} π r^{3} \times N_{A}}{135 \times 60}$

$= \frac{\frac{4}{3} π \times {(7.5 \sqrt{3} \times 10^{- 8})}^{3} \times 10^{23}}{135 \times 60} = 68 %$

$\Rightarrow$ unit cell is body centered

$\sqrt{3} a = 4 r$

$a = 4 \times 7.5 \times \frac{\sqrt{3}}{\sqrt{3}} = 30 Å = 3 nm$

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